AB Calculus (AP)

If s(x) = x and t(x) = x^2 +5x +6, then the function (s/t)(x) is completely described by: A. (s/t)(x) = x/x^2 +5x+6, restricted to the domain {x : x ≠ -2, x≠-3}

A. I'm lazy and there's too much to write, so there's the answer for you.

The domain of h(x) = {2x +1 x<1 {3 x≥1 is: A. All real numbers B. (-∞, 3) C. (-∞, 1) D. (-∞, 3] E. (-∞, 1]

A.

Consider the following functions: f(x) = cos(x^3 -x) h(x) = |x -3|^3 g(x) = ln(|x| +3) s(x) = sin^3(x) Which of the following is true? A. f and g are even, s is odd B. g and f are even, h is odd C. f is even, h and s are odd D. s is odd, f and h are even E. h and s are odd, g is even

A. First off, h can't be odd since h(0) ≠ 0. It also cannot be even since h(-1) = 64 ≠ 8 = h(1) Check on desmos.com/calculator for f and g, which are even, and s - which is odd.

Which of the following functions described does not have an inverse: A. A(t) = the altitude of an airplane as a function of time t between take-off and landing B. V(m) = the volume of m kilograms of sand C. A(d) = your age as a function of the date d D. K(t) = the temperature in Kelvin as a function of the temperature in t Fahrenheit E. None of these - they all have inverses

A. It is the only function here that is one-to-one!

If f(x) is an EVEN function and g(x) is an ODD function, which of the following must be EVEN? I. f(g(x)) II. f(x) + g(x) III. f(x)g(x) A. 1 only B. 2 only C. 1 & 2 only D. 2 & 3 only E. 1, 2, & 3

A. Look up flashcard #1 and #2 for the reason and use your head.

If g(x) = sinx on the domain [-π/2, π/2], the domain of g^-1 is: A. [-1, 1] B. [0,1] C. all real numbers D. [-√2/2, √2/2] E. [-π/2, π/2]

A. Range is E, Domain is A.

If h is the function given by h(x) = (f • g)(x) where f(x) = √x and g(x) = (√x)^3, then h(x) = A. x^2 restricted to the domain x ≥ 0 B. √x C. x^3 D. x^2 E. x^3/4 restricted to the domain x ≥ 0

A. h(x) = (f • g)(x) = f(x) • g(x) = √x •(√x)^3 = x^1/2 • x^3/2 = x^2

Suppose the following about a certain object: I. The object's altitude is a function of time, represented by the function A(x) where x is time II. The temperature of the object is a function of its velocity, represented by the function T(x), where here x is the object's velocity III. The weight of the object is a function of its altitude, represented by the function W(x), where here x is the object's altitude IV. The velocity of the object is a function of its weight, represented by the function V(x), where here x is the object's weight The function F(t) which gives us the object's temperature at time t can be represented as: A. F(t) = (T o W o A o V)(t) B. F(t) = (T o V o W o V)(t) C. F(t) = (A o W o V o T)(t) D. F(t) = (T o V o A o W)(t) E. F(t) = (T o W o V o A)(t)

B. B. F(t) = (T o V o W o V)(t) = (T(V(W(A(T))))) Time is input into the function A, which gives us altitude; the altitude is input into the function W, which gives is weight; this weight is input into the function V, which gives us velocity; and the velocity is input into the function T, which gives us temperature

Suppose you're given the following table of values for the function f(x), and you're told that the function is EVEN: | x || -2 || -.035 || 0 || 0.53 || 1 | |f(x)|| 5 || -3 || 2 || 2 || -5 | Then: A. f(2) = 5 B. f(-1) -f(2) = -10 C. Something is wrong. Given the table of values, the function can't be odd. D. f(0.35) +f(-0.53) = -1 E. f(0) +f(-0.53) = 0

B. Since the function is EVEN, you have E. f(0.35) +f(-0.53) = f(-0.35) +f(0.53) = -3 +2 = -1; 1 ≠ -1 D. f(0) +f(-0.53) = f(0) +f(0.53) = 2 +2 = 4; 4 ≠ 0 A. f(2) = f(-2) = 5; 5 ≠ -5 While B has f(-1) -f(2) = f(1) -f(-2) = -5 -5 = -10; -10 = -10

If f(x) is an odd function, which of the following must also be odd? A. |f(x)| B. -f(x) C. f(|x|) D. None of these E. f(x-1)

B. The function -f(x) is odd because -f(-x) = f(x) = -(-f(x)) A and C might be an even (don't quote me on that), but because of the subtraction in E, it is not an even or odd.

Suppose that f is an EVEN function whose domain is the set of all real numbers. Then which of the following can we claim to be true? A. The function has an inverse f^-1 that is odd. B. f^-1 is not a function C. We can't tell whether or not f has an inverse D. The function f has an inverse f^-1, but we can't tell whether it's even or odd. E. The function f has an inverse f^-1 that is even.

B. The inverse of f does not exist. Since f(-x) = f(x), the function is not one-to-one. For an inverse to exist, there must be a one-to-one function in the first place.

Given that f(x) = { x^3 if x ≥ 0 {x if x < 0 which of the following functions is even? I. f(x) II. f(|x|) III. |f(x)| A. 1 only B. 2 only C. 1 and 2 only D. 1 and 3 only E. None of these.

B. 1 is literally the function given in the beginning and that is NOT an even function, so literally cross out all except B and E. f(|x|) is even because f(|-x|) = f(|x|) but f(-x) ≠ f(x) and |f(-x)| ≠ |f(x)| so all except 2 are not even.

Suppose that f is a one-to-one function, and f^-1 is its inverse. Suppose also that h(x) = 4 and g(x) = x^2 +xsecx. Thien which of the following do we know NOT to be true? A. (g o h o f^-1)(x) = 16 +4sec4 B. (f o g o f^-1)(x) = x^2 +xsecx C. (h •(f^-1 o f^-1 o f o f o g))(x) = 4x^2 +4xsecx D. (f o f^-1 oh)(x) = 4 E. (h o g o g)(x) = 4

B. We don't know that (f o g o f^-1)(x) = x^2 +xsecx since we don't know what the rules for f and f^-1 are. Also because f and f^-1 are separated by that g. E. (h o g o g)(x) = h(g(g(x))) = h(g(x^2 +xsecx)) = 4 since there isn't an x for g(g(x)) to insert into h(x) D. (f o f^-1 o h)(x) = f(f^-1(h(x))) = f(f^-1(4)) = 4; C. (g o h o f^-1)(x) = g(h(f^-1(x))) = g(4) = 16+4sec4 A. h(x)•g(x) since f and f^-1 cancel out = 4x^2 +4xsecx, which is obviously incorrect

If f(x) is an ODD function and g(x) is an ODD function, which of the following must be ODD? I. f(g(x)) II. f(x) + g(x) III. f(x)g(x) A. 1 only B. 2 only C. 1 & 2 only D. 2 & 3 only E. 1, 2, & 3

C. Look up flashcards #1 and #2 for the reason and use your head. I. f(g(-x)) = f(-g(x)) = -f(g(x)); Odd II. f(-x) +g(-1) = -f(x) -g(x) = -(f(x) +g(x)); Odd III. f(-x)g(-x) = -f(x))(-g(x)) = f(x)g(x); Even

Consider the following functions: f(x) = sin(x^4-x^2) h(x) = (|x| -3)^3 g(x) = ln(|x|) +3 s(x) = sin^3(x) Which of the following is true? (Hint: use desmos.com/calculator to graph) A. f, h, and s are odd B. f is even, h and s are odd C. f and h are even, s is odd D. h and g are even, f and s are odd E. h and s are even, f is odd

C. Okay, process of elimination: f is even, so that immediately crosses out A, D, and E. h(x) is the determining factor - oh, it's even!

If g(x) = x/x+1 then the inverse function g^-1(x) = A. x/x+1 B. x-1/x C. x/1-x D. x+1/x E. x

C. Plug it in.

Suppose that f is an ODD function whose domain is the set of all real numbers. Then which of the following can we claim to be true? A. The function f has an inverse f^-1 that is ODD B. The function f does not have an inverse C. We can't tell whether f has an inverse that's still a function. D. The function f has an inverse f^-1 that is EVEN E. The function f has an inverse f^-1, but we can't tell whether it's even or odd

C. The function f(x) = x is odd and has an inverse. The function f(x) = sinx is odd and does not have an inverse.

Suppose you're given the following table of values for the function f(x), and you're told that the function is ODD: | x || -2 || -.035 || 0 || 0.53 || 1 | |f(x)|| 5 || -3 || 2 || 2 || -5 | Then: A. f(2) = 5 B. f(-1) -f(2) = -10 C. Something is wrong. Given the table of values, the function can't be odd. D. f(0.35) +f(-0.53) = -1 E. f(0) +f(-0.53) = 0

C. Given the table, the function can't be odd. An odd function must ALWAYS have f(0) = 0 since we need f(0) = -f(-0) = -f(0)! Here, we have f(0) = 2, which is obviously wrong. In other words, an ODD function must always cross the origin.

What is the domain of the function f given by f(x) = √(x^2 -4)/x-3? A. {x:x≠3} B. {x: |x| ≤ 2} C. {x: |x| ≥ 2} D. {x: |x| ≥ 2 and x ≠3} E. {x: x ≥ 2 and x ≠ 3}

D.

If f(x) is an odd function, which of the following must be even? A. f(|x-1|) B. None of these C. f(x +1) D. |f(x)| E. -f(x)

D. Right off the bat, there are two suspicious ones. A because of the subtraction and C because of the addition. E is an odd function, so D |f(-x)| = |-f(x)| = |f(x)|

Suppose F(x) = x^2/x^2 -7, and F(x) = (h o g)(x). Then which of the following are possible definitions for h and g? I. h(x) = x, g(x) = x^2/x^2 -7 II. h(x) = x^2, g(x) = x/x-7 III. h(x) = x^4/x^4-7, g(x) = √|x| IV. h(x) = x/x-7, g(x) = x^2 A. 2 & 3 only B. 1 & 2 only C. 1 & 4 only D. 1, 3, & 4 only E. 1, 2, 3, & 4

D. If h(x) = x^2, g(x) = x/x-7, then (h o g)(x) = h(g(x)) = h(x/x-7) = (x/x-7)^2 ≠ F(x)

f(x) = [[x]] is the greatest integer function. The range of f(x) = x -[[x]] is: A. (0, 1) B. All real numbers C. (0,1] D. [0, 1) E. [0, 1]

E. Similar to Question 20

The range of g(x) = [[x/2]] A. All even integers B. All negative integers C. All odd integers D. All real numbers E. All integers

E. [[]] represents the greatest integer That is why it's always 'integers'. This means all integers.

If h is the function given by h(x) = (f o g)(x), where f(x) = 5x^2 and g(x) = |x|, then h(x) = A. |5x^2 -1| B. 5x^2 -|x| C. 5/x| -1 D. 5x -1 E. 5|x|^2 -1

E. h(x) = (f o g)(x) = f(g(x)) = f(|x|) = 5(|x|)^2 -1 = 5|x|^2 -1

True or False: If f(x) and g(x) are both functions of x, then domain of (f•g)(x) = domain of f(x) U domain of g(x)

False.

Regarding algebra of functions (+, -, •, ÷): • The sum of two even functions is even. • The sum of two odd functions is odd. • The difference of two even functions is even. • The difference of two odd functions is odd. • The product of two even functions is even. • The product of two odd functions is even. • The quotient of two even functions is even. • The quotient of two odd functions is even.

Regarding composition of functions ((f o g)(x)): • The composition of two EVEN functions is EVEN. • The composition of two ODD functions is ODD. • The composition of an EVEN function and an ODD function is EVEN.

Properties of Odd and Even Functions: The only function that is both odd AND even is f(x) = 0 If a function is ODD, the absolute value of that function is EVEN.

Visit https://mathbitsnotebook.com/Algebra2/Functions/FNOddEvenFunctions.html for more information.